Question

If a tangent to the hyperbola \( xy = -1 \) is also a tangent to the parabola \( y^2 = 8x \), then the equation of that tangent is:

• A. \( 3y + x = 2 \)
• B. \( y = 3x + 4 \)
• C. \( y = x + 2 \)
• D. \( y = 2x + 1 \)

Hint:

To check if a line is a tangent to a conic, substitute the line into the conic and check whether the resulting equation has exactly one solution (discriminant = 0).

Answer 1

The Correct Option is C

Step 1: General form of tangent to hyperbola 
For the hyperbola \( xy = -1 \), the general tangent is: 
\[ y = mx + \frac{-1}{m} \] This line is a tangent to the parabola \( y^2 = 8x \) as well. 
Step 2: Use condition for tangent to parabola 
The condition for the line \( y = mx + c \) to be a tangent to the parabola \( y^2 = 8x \) is: 
\[ c = \frac{2}{m} \] Step 3: Equate both expressions for the constant term 
From the hyperbola's tangent, \( c = \frac{-1}{m} \) 
From the parabola's tangent, \( c = \frac{2}{m} \) 
\[ \frac{-1}{m} = \frac{2}{m} \Rightarrow -1 = 2 \] This is not valid. So instead, plug \( y = mx + \frac{-1}{m} \) into the parabola \( y^2 = 8x \) and demand the resulting quadratic has exactly one solution (discriminant zero). 
Step 4: Substitute into parabola and solve 
\[ \left(mx + \frac{-1}{m}\right)^2 = 8x \Rightarrow m^2x^2 - 2x + \frac{1}{m^2} = 8x \Rightarrow m^2x^2 - 10x + \frac{1}{m^2} = 0 \] This is a quadratic in \( x \). For the line to be tangent to the parabola, the discriminant must be zero: \[ \text{Discriminant } D = (-10)^2 - 4(m^2)\left(\frac{1}{m^2}\right) = 100 - 4 = 96 \neq 0 \] Let us now try the options directly. 
Step 5: Check each option 
Option (3): \( y = x + 2 \) 
- For hyperbola \( xy = -1 \): Substitute \( y = x + 2 \Rightarrow x(x + 2) = -1 \Rightarrow x^2 + 2x + 1 = 0 \Rightarrow (x + 1)^2 = 0 \) Only one solution ⇒ it is a tangent. 
- For parabola \( y^2 = 8x \): Substitute \( y = x + 2 \Rightarrow (x + 2)^2 = 8x \Rightarrow x^2 + 4x + 4 = 8x \Rightarrow x^2 - 4x + 4 = 0 \Rightarrow (x - 2)^2 = 0 \) 
Again, one solution ⇒ it is a tangent. 
Hence, \( \boxed{y = x + 2} \) is a common tangent.